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A student measures the depth of a water well with an adjustable frequency audio oscillator. 2 successive resonant frequencies are heard at 4
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Answers ( )
Answer:
17.15 m
Explanation:
Assuming the well is empty (full of air instead of water), the speed of sound is v = 343 m/s.
The water well acts as a pipe closed at one end. Therefore, the depth of the water well must be an odd multiple of a quarter of the resonant wavelength.
L = (2n − 1) λ/4, n = 1, 2, 3, etc.
v = λf, so λ = v/f. Substituting:
L = (2n − 1) v/(4f)
Solving for frequency:
f = (2n − 1) v/(4L)
The difference between two successive resonant frequencies is therefore:
Δf = (2(n+1) − 1) v/(4L) − (2n − 1) v/(4L)
Δf = (2n + 1) v/(4L) − (2n − 1) v/(4L)
Δf = 2 v/(4L)
Δf = v/(2L)
Plugging in values:
50 Hz − 40 Hz = 343 m/s / (2L)
2L = 34.3 m
L = 17.15 m